Answer:
<em>The rise in temperature will be less than 5 °C in the second minute.</em>
Explanation:
According to heat conduction law, the rate of heating is proportional to the temperature difference or temperature gradient. The temperature gradient is what drives heat to move from a hotter body at a higher temperature gradient to a colder body at a lower temperature gradient. For the potato, the initial first minute raises the temperature to 5 °C, consequently reducing the temperature gradient between the potato and the heating element in the oven. <em>This reduced temperature gradient means that the rate at which it will conduct heat in the second minute will be lesser than that at the first minute</em>. This will continue till the potato and the heating element are at the same temperature, at which no temperature gradient will exist between them; stopping heat transfer between them.
Answer:
Q = 14.578 m³/s
Explanation:
Given
We use the Manning Equation as follows
Q = (1/n)*A*(∛R²)*(√S)
where
- Q = volumetric water flow rate passing through the stretch of channel (m³/s for S.I.)
-
A = cross-sectional area of flow perpendicular to the flow direction, (m² for S.I.)
-
S = bottom slope of channel, m/m (dimensionless) = 2.5% = 0.025
-
n = Manning roughness coefficient (empirical constant), dimensionless = 0.023
-
R = hydraulic radius = A/P (m for S.I.) where
:
-
A = cross-sectional area of flow as defined above,
-
P = wetted perimeter of cross-sectional flow area (m for S.I.)
we get A as follows
A = (B*h)/2
where
B = 5 m (the top width of the flowing channel)
h = (B/2)*(m) = (5 m/2)*(1/2) = 1.25 m (the deep)
A = (5 m*1.25 m/2) = 3.125 m²
then we find P
P = 2*√((B/2)²+h²) ⇒ P = 2*√((2.5 m)²+(1.25 m)²) = 5.59 m
⇒ R = A/P ⇒ R = 3.125 m²/5.59 m = 0.559 m
Substituting values into the Manning equation gives:
Q = (1/0.023)*(3.125 m²)*(∛(0.559 m)²)*(√0.025)
⇒ Q = 14.578 m³/s
Search up A gardener can increase the number of dahlia plants in an annual garden by either buying new bulbs each year or dividing the existing bulbs to create new plants . The table below shows the expected number of bulbs for each method
Part A
For each method,a function to model the expected number of plants for each year
Part B
Use the Functions to Find the expected number of plants in 10 years for each method.
Part C
3-SAT ≤p TSP
If P ¹ NP, then no NP-complete problem can be solved in polynomial time.
both the statements are true.
<u>Explanation:</u>
- 3-SAT ≤p TSP due to any complete problem of NP to other problem by exits of reductions.
- If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
- If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.
Answer:
The required pumping head is 1344.55 m and the pumping power is 236.96 kW
Explanation:
The energy equation is equal to:
For the pipe 1, the flow velocity is:
Q = 18 L/s = 0.018 m³/s
D = 6 cm = 0.06 m
The Reynold´s number is:
Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941
The head of pipe 1 is:
For the pipe 2, the flow velocity is:
The Reynold´s number is:
The head of pipe 1 is:
The total head is:
hi = 1326.18 + 21.3 = 1347.48 m
The required pump head is:
The required pumping power is: